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This is a question that I read,

and I am curious about how to solve $au_x+bu_y+cu=0$ by interpreting geometrically.

Additionally, the second answer on that page says

Let $v(x,y)=e^{cx}u(x,y)$ and compute $v_x,v_y$ ...

but how can we think up multiplying $e^{cx}$ in to $u(x,y)$ ?

stoneaa
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1 Answers1

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I would try "separation" of variables. Let u(x,y)= X(x)Y(y). That is, look for a solution that is a function of one function of x only and a function of y only. $u_x= X'Y$ and $u_y= XY'$ so the equation is $aX'Y+ bXY'+ cXY= 0$. Divide by u= XY: $a\frac{X'}{X}+ b\frac{Y'}{Y}+ 1= 0$. That must be true for all x and y. In particular if we hold y constant, so that Y'= 0, and vary only x, $b\frac{X'}{X}+ 1= 0$. If we hold x constant, so that X'= 0, and vary only y, $a\frac{Y'}{Y}+ 1= 0$. That means that $a\frac{dX}{dx}$ and $b\frac{dY}{Y}$ must be constants. We must have $a\frac{dX}{dx}= A$ and $b\frac{dY}{.dy}= B$ with A+ B= 1.

user247327
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